Examples

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦

$A = [\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}]$

Step 1

Set up the formula to find the characteristic equation

p (λ)

$p (λ)$ .

p (λ) = determinant (A - λ I_{3})

$p (λ) = determinant (A - λ I_{3})$

Step 2

The identity matrix or unit matrix of size

3

$3$ is the

3 \times 3

$3 \times 3$ square matrix with ones on the main diagonal and zeros elsewhere.

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 3

Substitute the known values into

p (λ) = determinant (A - λ I_{3})

$p (λ) = determinant (A - λ I_{3})$ .

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Step 3.1

Substitute

⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}]$ for

A

$A$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ - λ I_{3} ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] - λ I_{3})$

Step 3.2

Substitute

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ for

I_{3}

$I_{3}$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ - λ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ - λ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply

- λ

$- λ$ by each element of the matrix.

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & - λ \cdot 0 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 λ & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

Step 4.1.2.3

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 λ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

Step 4.1.2.4

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 λ & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

Step 4.1.2.5

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 λ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 & - λ \cdot 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 & 0 λ & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 & 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 & 0 & - λ \cdot 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 & 0 & - λ \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 & 0 & - λ \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 1 1 & 0 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - λ & 0 & 0 0 & - λ & 0 0 & 0 & - λ \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

$p (λ) = determinant ([\begin{array}{c} 2 & 2 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 + 0 & 1 + 0 1 + 0 & 0 - λ & 0 + 0 0 + 0 & 2 + 0 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 + 0 & 1 + 0 \\ 1 + 0 & 0 - λ & 0 + 0 \\ 0 + 0 & 2 + 0 & 1 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add

2

$2$ and

0

$0$ .

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 + 0 1 + 0 & 0 - λ & 0 + 0 0 + 0 & 2 + 0 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 + 0 \\ 1 + 0 & 0 - λ & 0 + 0 \\ 0 + 0 & 2 + 0 & 1 - λ \end{array}]$

Step 4.3.2

Add

1

$1$ and

0

$0$ .

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 1 + 0 & 0 - λ & 0 + 0 0 + 0 & 2 + 0 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 \\ 1 + 0 & 0 - λ & 0 + 0 \\ 0 + 0 & 2 + 0 & 1 - λ \end{array}]$

Step 4.3.3

Add

1

$1$ and

0

$0$ .

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 1 & 0 - λ & 0 + 0 0 + 0 & 2 + 0 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 \\ 1 & 0 - λ & 0 + 0 \\ 0 + 0 & 2 + 0 & 1 - λ \end{array}]$

Step 4.3.4

Subtract

λ

$λ$ from

0

$0$ .

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 1 & - λ & 0 + 0 0 + 0 & 2 + 0 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 \\ 1 & - λ & 0 + 0 \\ 0 + 0 & 2 + 0 & 1 - λ \end{array}]$

Step 4.3.5

Add

0

$0$ and

0

$0$ .

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 1 & - λ & 0 0 + 0 & 2 + 0 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 \\ 1 & - λ & 0 \\ 0 + 0 & 2 + 0 & 1 - λ \end{array}]$

Step 4.3.6

Add

0

$0$ and

0

$0$ .

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 1 & - λ & 0 0 & 2 + 0 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 \\ 1 & - λ & 0 \\ 0 & 2 + 0 & 1 - λ \end{array}]$

Step 4.3.7

Add

2

$2$ and

0

$0$ .

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 1 & - λ & 0 0 & 2 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 \\ 1 & - λ & 0 \\ 0 & 2 & 1 - λ \end{array}]$

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 1 & - λ & 0 0 & 2 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 \\ 1 & - λ & 0 \\ 0 & 2 & 1 - λ \end{array}]$

p (λ) = determinant ⎡ ⎢ ⎣ \begin{matrix} 2 - λ & 2 & 1 1 & - λ & 0 0 & 2 & 1 - λ \end{matrix} ⎤ ⎥ ⎦

$p (λ) = determinant [\begin{array}{c} 2 - λ & 2 & 1 \\ 1 & - λ & 0 \\ 0 & 2 & 1 - λ \end{array}]$

Step 5

Find the determinant.

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Step 5.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in column

1

$1$ by its cofactor and add.

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Step 5.1.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

Step 5.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} - λ & 0 2 & 1 - λ \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - λ & 0 \\ 2 & 1 - λ \end{array} |$

Step 5.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

(2 - λ) ∣ ∣ ∣ \begin{matrix} - λ & 0 2 & 1 - λ \end{matrix} ∣ ∣ ∣

$(2 - λ) | \begin{array}{c} - λ & 0 \\ 2 & 1 - λ \end{array} |$

Step 5.1.5

The minor for

a_{21}

$a_{21}$ is the determinant with row

2

$2$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} |$

Step 5.1.6

Multiply element

a_{21}

$a_{21}$ by its cofactor.

- 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣

$- 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} |$

Step 5.1.7

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 2 & 1 - λ & 0 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 2 & 1 \\ - λ & 0 \end{array} |$

Step 5.1.8

Multiply element

a_{31}

$a_{31}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 2 & 1 - λ & 0 \end{matrix} ∣ ∣ ∣

$0 | \begin{array}{c} 2 & 1 \\ - λ & 0 \end{array} |$

Step 5.1.9

Add the terms together.

p (λ) = (2 - λ) ∣ ∣ ∣ \begin{matrix} - λ & 0 2 & 1 - λ \end{matrix} ∣ ∣ ∣ - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 2 & 1 - λ & 0 \end{matrix} ∣ ∣ ∣

$p (λ) = (2 - λ) | \begin{array}{c} - λ & 0 \\ 2 & 1 - λ \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0 | \begin{array}{c} 2 & 1 \\ - λ & 0 \end{array} |$

p (λ) = (2 - λ) ∣ ∣ ∣ \begin{matrix} - λ & 0 2 & 1 - λ \end{matrix} ∣ ∣ ∣ - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 2 & 1 - λ & 0 \end{matrix} ∣ ∣ ∣

$p (λ) = (2 - λ) | \begin{array}{c} - λ & 0 \\ 2 & 1 - λ \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0 | \begin{array}{c} 2 & 1 \\ - λ & 0 \end{array} |$

Step 5.2

Multiply

0

$0$ by

∣ ∣ ∣ \begin{matrix} 2 & 1 - λ & 0 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 2 & 1 \\ - λ & 0 \end{array} |$ .

p (λ) = (2 - λ) ∣ ∣ ∣ \begin{matrix} - λ & 0 2 & 1 - λ \end{matrix} ∣ ∣ ∣ - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) | \begin{array}{c} - λ & 0 \\ 2 & 1 - λ \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3

Evaluate

∣ ∣ ∣ \begin{matrix} - λ & 0 2 & 1 - λ \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - λ & 0 \\ 2 & 1 - λ \end{array} |$ .

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Step 5.3.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

p (λ) = (2 - λ) (- λ (1 - λ) - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ (1 - λ) - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2

Simplify the determinant.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Apply the distributive property.

p (λ) = (2 - λ) (- λ \cdot 1 - λ (- λ) - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ \cdot 1 - λ (- λ) - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.1.2

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = (2 - λ) (- λ - λ (- λ) - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ - λ (- λ) - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.1.3

Rewrite using the commutative property of multiplication.

p (λ) = (2 - λ) (- λ - 1 \cdot - 1 λ \cdot λ - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ - 1 \cdot - 1 λ \cdot λ - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.1.4

Simplify each term.

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Step 5.3.2.1.4.1

Multiply

λ

$λ$ by

λ

$λ$ by adding the exponents.

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Step 5.3.2.1.4.1.1

Move

λ

$λ$ .

p (λ) = (2 - λ) (- λ - 1 \cdot - 1 (λ \cdot λ) - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ - 1 \cdot - 1 (λ \cdot λ) - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.1.4.1.2

Multiply

λ

$λ$ by

λ

$λ$ .

p (λ) = (2 - λ) (- λ - 1 \cdot - 1 λ^{2} - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ - 1 \cdot - 1 λ^{2} - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

p (λ) = (2 - λ) (- λ - 1 \cdot - 1 λ^{2} - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ - 1 \cdot - 1 λ^{2} - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.1.4.2

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

p (λ) = (2 - λ) (- λ + 1 λ^{2} - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ + 1 λ^{2} - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.1.4.3

Multiply

λ^{2}

$λ^{2}$ by

1

$1$ .

p (λ) = (2 - λ) (- λ + λ^{2} - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ + λ^{2} - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

p (λ) = (2 - λ) (- λ + λ^{2} - 2 \cdot 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ + λ^{2} - 2 \cdot 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.1.5

Multiply

- 2

$- 2$ by

0

$0$ .

p (λ) = (2 - λ) (- λ + λ^{2} + 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ + λ^{2} + 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

p (λ) = (2 - λ) (- λ + λ^{2} + 0) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ + λ^{2} + 0) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.2

Add

- λ + λ^{2}

$- λ + λ^{2}$ and

0

$0$ .

p (λ) = (2 - λ) (- λ + λ^{2}) - 1 ∣ ∣ ∣ \begin{matrix} 2 & 1 2 & 1 - λ \end{matrix} ∣ ∣ ∣ + 0

$p (λ) = (2 - λ) (- λ + λ^{2}) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.3.2.3

Reorder

$- λ$ and

$λ^{2}$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 | \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} | + 0$

Step 5.4

Evaluate

$| \begin{array}{c} 2 & 1 \\ 2 & 1 - λ \end{array} |$ .

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Step 5.4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 (1 - λ) - 2 \cdot 1) + 0$

Step 5.4.2

Simplify the determinant.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Apply the distributive property.

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 \cdot 1 + 2 (- λ) - 2 \cdot 1) + 0$

Step 5.4.2.1.2

Multiply

$2$ by

$1$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 + 2 (- λ) - 2 \cdot 1) + 0$

Step 5.4.2.1.3

Multiply

$- 1$ by

$2$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 - 2 λ - 2 \cdot 1) + 0$

Step 5.4.2.1.4

Multiply

$- 2$ by

$1$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 - 2 λ - 2) + 0$

Step 5.4.2.2

Combine the opposite terms in

$2 - 2 λ - 2$ .

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Step 5.4.2.2.1

Subtract

$2$ from

$2$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (- 2 λ + 0) + 0$

Step 5.4.2.2.2

Add

$- 2 λ$ and

$0$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (- 2 λ) + 0$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Add

$(2 - λ) (λ^{2} - λ) - 1 (- 2 λ)$ and

$0$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (- 2 λ)$

Step 5.5.2

Simplify each term.

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Step 5.5.2.1

Expand

$(2 - λ) (λ^{2} - λ)$ using the FOIL Method.

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Step 5.5.2.1.1

Apply the distributive property.

$p (λ) = 2 (λ^{2} - λ) - λ (λ^{2} - λ) - 1 (- 2 λ)$

Step 5.5.2.1.2

Apply the distributive property.

$p (λ) = 2 λ^{2} + 2 (- λ) - λ (λ^{2} - λ) - 1 (- 2 λ)$

Step 5.5.2.1.3

Apply the distributive property.

$p (λ) = 2 λ^{2} + 2 (- λ) - λ \cdot λ^{2} - λ (- λ) - 1 (- 2 λ)$

Step 5.5.2.2

Simplify and combine like terms.

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Step 5.5.2.2.1

Simplify each term.

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Step 5.5.2.2.1.1

Multiply

$- 1$ by

$2$ .

$p (λ) = 2 λ^{2} - 2 λ - λ \cdot λ^{2} - λ (- λ) - 1 (- 2 λ)$

Step 5.5.2.2.1.2

Multiply

$λ$ by

$λ^{2}$ by adding the exponents.

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Step 5.5.2.2.1.2.1

Move

$λ^{2}$ .

$p (λ) = 2 λ^{2} - 2 λ - (λ^{2} λ) - λ (- λ) - 1 (- 2 λ)$

Step 5.5.2.2.1.2.2

Multiply

$λ^{2}$ by

$λ$ .

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Step 5.5.2.2.1.2.2.1

Raise

$λ$ to the power of

$1$ .

$p (λ) = 2 λ^{2} - 2 λ - (λ^{2} λ^{1}) - λ (- λ) - 1 (- 2 λ)$

Step 5.5.2.2.1.2.2.2

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = 2 λ^{2} - 2 λ - λ^{2 + 1} - λ (- λ) - 1 (- 2 λ)$

Step 5.5.2.2.1.2.3

Add

$2$ and

$1$ .

$p (λ) = 2 λ^{2} - 2 λ - λ^{3} - λ (- λ) - 1 (- 2 λ)$

Step 5.5.2.2.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = 2 λ^{2} - 2 λ - λ^{3} - 1 \cdot - 1 λ \cdot λ - 1 (- 2 λ)$

Step 5.5.2.2.1.4

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 5.5.2.2.1.4.1

Move

$λ$ .

$p (λ) = 2 λ^{2} - 2 λ - λ^{3} - 1 \cdot - 1 (λ \cdot λ) - 1 (- 2 λ)$

Step 5.5.2.2.1.4.2

Multiply

$λ$ by

$λ$ .

$p (λ) = 2 λ^{2} - 2 λ - λ^{3} - 1 \cdot - 1 λ^{2} - 1 (- 2 λ)$

Step 5.5.2.2.1.5

Multiply

$- 1$ by

$- 1$ .

$p (λ) = 2 λ^{2} - 2 λ - λ^{3} + 1 λ^{2} - 1 (- 2 λ)$

Step 5.5.2.2.1.6

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = 2 λ^{2} - 2 λ - λ^{3} + λ^{2} - 1 (- 2 λ)$

Step 5.5.2.2.2

Add

$2 λ^{2}$ and

$λ^{2}$ .

$p (λ) = 3 λ^{2} - 2 λ - λ^{3} - 1 (- 2 λ)$

Step 5.5.2.3

Multiply

$- 2$ by

$- 1$ .

$p (λ) = 3 λ^{2} - 2 λ - λ^{3} + 2 λ$

Step 5.5.3

Combine the opposite terms in

$3 λ^{2} - 2 λ - λ^{3} + 2 λ$ .

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Step 5.5.3.1

Add

$- 2 λ$ and

$2 λ$ .

$p (λ) = 3 λ^{2} - λ^{3} + 0$

Step 5.5.3.2

Add

$3 λ^{2} - λ^{3}$ and

$0$ .

$p (λ) = 3 λ^{2} - λ^{3}$

Step 5.5.4

Reorder

$3 λ^{2}$ and

$- λ^{3}$ .

$p (λ) = - λ^{3} + 3 λ^{2}$

Step 6

Set the characteristic polynomial equal to

$0$ to find the eigenvalues

$λ$ .

$- λ^{3} + 3 λ^{2} = 0$

Step 7

Solve for

$λ$ .

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Step 7.1

Factor

$- λ^{2}$ out of

$- λ^{3} + 3 λ^{2}$ .

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Step 7.1.1

Factor

$- λ^{2}$ out of

$- λ^{3}$ .

$- λ^{2} λ + 3 λ^{2} = 0$

Step 7.1.2

Factor

$- λ^{2}$ out of

$3 λ^{2}$ .

$- λ^{2} λ - λ^{2} \cdot - 3 = 0$

Step 7.1.3

Factor

$- λ^{2}$ out of

$- λ^{2} (λ) - λ^{2} (- 3)$ .

$- λ^{2} (λ - 3) = 0$

Step 7.2

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$λ^{2} = 0$

$λ - 3 = 0$

Step 7.3

Set

$λ^{2}$ equal to

$0$ and solve for

$λ$ .

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Step 7.3.1

Set

$λ^{2}$ equal to

$0$ .

$λ^{2} = 0$

Step 7.3.2

Solve

$λ^{2} = 0$ for

$λ$ .

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Step 7.3.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$λ = \pm \sqrt{0}$

Step 7.3.2.2

Simplify

$\pm \sqrt{0}$ .

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Step 7.3.2.2.1

Rewrite

$0$ as

$0^{2}$ .

$λ = \pm \sqrt{0^{2}}$

Step 7.3.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$λ = \pm 0$

Step 7.3.2.2.3

Plus or minus

$0$ is

$0$ .

$λ = 0$

Step 7.4

Set

$λ - 3$ equal to

$0$ and solve for

$λ$ .

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Step 7.4.1

Set

$λ - 3$ equal to

$0$ .

$λ - 3 = 0$

Step 7.4.2

Add

$3$ to both sides of the equation.

$λ = 3$

Step 7.5

The final solution is all the values that make

$- λ^{2} (λ - 3) = 0$ true.

$λ = 0, 3$

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